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I want to show that every bounded sequence has a subsequence that converge. Do you agree with my proof and if not, what's wrong ?

Proof

Let $(x_n)_{n\in\mathbb N^*}$ a bounded sequence. Since $(x_n)_{n\in\mathbb N^*}$ is bounded, $\limsup_{n\to\infty }(x_n)_{n\in\mathbb N^*}$ exist. Indeed, if we set for all $n$, $$y_n:=\sup_{k\geq n}x_k,$$ the sequence $(y_n)_{n\in\mathbb N^*}$ is decreasing and bounded end thus convergente. Let $$\ell:=\inf_{n\in\mathbb N}\sup_{k\geq n}x_k=\inf_{n\in\mathbb N} y_n.$$

Let $k\in\mathbb N^*$. By definition of $\ell$, $$\exists n_k\in\mathbb N: |y_{n_k}-\ell|<\frac{1}{2k}.$$

We can suppose with out loss of generality that $(n_k)_{k\in\mathbb N^*}$ is strictly increasing. Indeed, by convergence of $(y_{n_k})_{k\in\mathbb N^*}$, $$\exists n_{k+1}>n_k:\forall n\in\mathbb N, n\geq n_{k+1}\implies |y_{n_{k+1}}-\ell|<\frac{1}{2(k+1)}.$$

By definition of $y_{n_k}$, $$\exists m_k\geq k :|y_{n_{k}}-x_{n_{m_k}}|<\frac{1}{2k}.$$

Let show that $(n_{m_k})_{k\in\mathbb N^*}$ is strictly increasing. Since $(n_p)_{p\in\mathbb N^*}$ est strictly increasing, we just have to show that $(m_k)_{k\in\mathbb N^*}$ is strictly increasing too to conclude that $(n_{m_k})_{k\in\mathbb N^*}$ is strictly increasing. Suppose by contradiction that it's not strictly increasing, i.e. that $$\left[y_{n_{k+1}}-\frac{1}{2(k+1)}, y_{n_{k+1}}\right)\cap\{x_{n_p}\mid p>m_k\}=\emptyset.$$ If $m_k=k$, we have a contradiction with the existence of $y_{n_{k+1}}$, then we can suppose that $m_{k}\geq k+1$. Therefore, $$y_{n_{k+1}}=\max_{p=k+1,...,m_k}x_{n_p},$$ and thus, for all $q\geq 1$, $$y_{n_{k+q}}=\max_{p=k+q,...,m_k}x_{n_p}.$$ However, for all $q>k-m_{k}$, we have that $y_{n_q}$ doesn't exist, which is a contradiction with the fact that $(y_n)_{n\in\mathbb N^*}$ converge. Then, the sequence $(m_k)_{k\in\mathbb N^*}$ can be supposed strictly increasing.

To conclude,

$$|x_{n_{m_k}}-\ell|\leq |x_{n_{m_k}}-y_{n_{k}}|+|y_{n_{k}}-\ell|\leq\frac{1}{2k}+\frac{1}{2k}=\frac{1}{k}\underset{k\to\infty }{\longrightarrow }0,$$ and thus, $(x_{n_{m_k}})_{k\in\mathbb N^*}$ is a subsequence of $(x_n)_{n\in\mathbb N^*}$ that converge.

Gregory Grant
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idm
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  • Looks good. But I don't know what $\mathbb{N}^*$ is. – Michael May 15 '15 at 15:19
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    the use of the iterated subscript in $x_{n_{m_k}}$ is a neat way to handle the extraction of the convergent subsequence, having used the inf sup operator to select a suitable target for convergence. – David Holden May 15 '15 at 15:26
  • Actually it would reduce scary notation with triple subscripts if you just said "by definition of $y_{n_k}$, there is an $m_k \geq n_k$ such that $|y_{n_k}-x_{m_k}| \leq 1/(2k)$. But that is just a cosmetic issue. – Michael May 15 '15 at 15:30
  • @Michael: Yes, I corrected it, thanks :-) – idm May 15 '15 at 15:31
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    @idm Actually there is a problem with this proof. You have to make sure the sequence of natural numbers $n_{m_k}$ is strictly increasing. I don't think you guarantee that anywhere, do you? – Gregory Grant May 15 '15 at 15:31
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    @Gregory: You right, this was my problem actually. I'll think about that, thankd :-) – idm May 15 '15 at 15:32
  • @idm I went ahead and posted my comment and response to your follow-up comment, as an answer – Gregory Grant May 15 '15 at 15:35
  • @Michael: Actually it's $m_k\geq k$. Indeed, $y_{n_k}=\sup_{j\geq k}x_{n_j}$ – idm May 15 '15 at 15:46
  • I thought $y_{n_k} = \sup_{j \geq n_k} x_{j}$ (just applying the definition). If your other result is also true by some property of the $x_{n_j}$ sequence, it kind of puts a lot of burden on the reader to keep track of all these subscripts. If you want to make sure the indices diverge to infinity, why not just choose $n_k$ larger than $n_{k-1}$ and such that the desired property $|y_{n_k}-l|< 1/(2k)$ holds? – Michael May 15 '15 at 15:57
  • The fact that we can choose $n_k>n_{k-1}$ is a consequence of the convergence of $(y_n)$. I think that I have shown that $(n_{m_k})$ is strictly increasing. You can see the proof as a comment to the answer of Gregory Grant. If any body has time, tell me if you are satisfy. Thanks :-) – idm May 15 '15 at 19:05
  • Can you tell if this proof is correct :-) – idm May 16 '15 at 14:38
  • @idm I hope you don't mind I changed your interval notation to be more standard. I'll try to find a minute later to check the logic. – Gregory Grant May 16 '15 at 14:41
  • @G.Grant: Thanks a lot :-) – idm May 16 '15 at 15:19

1 Answers1

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Actually there is a problem with this proof. You have to make sure the sequence of natural numbers $n_{m_k}$ is strictly increasing. I don't think you guarantee that anywhere, do you?

That's why you have to define $n_{m_k}$ iteratively, since at each step there are infinitely many subscripts that satisfy the condition, you can always find one further out than the previous.

Otherwise it's a real nice proof, nice and concise. I wonder, however, if you can't simplify it further, because you never actually use that $y_n$ converges. If you do that maybe you can avoid the triple subscript.

Gregory Grant
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